Free FE Civil Practice Problems with Step-by-Step Solutions

Step 1 – Identify given information: Warren-type truss with \(12\,\mathrm{kN}\) load at B. At joint D: members AD and DC are horizontal (collinear), BD is vertical. No external load at D.

Step 2 – Zero-force member identification (FE Handbook): At a joint where two members are collinear and a third member is non-collinear, if there is no external load at that joint, the non-collinear member is a zero-force member.

Step 3 – Apply at joint D:

AD and DC are both horizontal (collinear). BD is vertical (non-collinear). No external load acts at D.

$$\sum F_y = 0: \quad F_{BD} = 0$$

Step 4 – Answer: \(F_{BD} = 0\,\mathrm{kN}\). BD is a zero-force member. Options (B), (C), and (D) all assign nonzero forces, which would violate vertical equilibrium at joint D. Zero-force members provide stability but carry no load under this loading.

Correct answer: (A)

Step 1 – Identify given information:

\( L = 6 \;\text{m} \), two equal loads of \( 4 \;\text{kN} \) at \( x = 2 \;\text{m} \) and \( x = 4 \;\text{m} \).

Step 2 – Use symmetry and statics (FE Handbook):

By symmetry: \( R_A = R_B = 4 \;\text{kN} \).

Step 3 – Solve:

Between the loads, shear is \( V = R_A - 4 = 0 \), so moment is constant there:

$$M_{\max} = R_A \times 2 = 4 \times 2 = 8 \;\text{kN}\cdot\text{m}$$

Shear (V) and bending-moment (M) diagrams

Step 4 – Answer: The maximum moment is \( 8 \;\text{kN}\cdot\text{m} \), constant between the two loads. This is a classic four-point bending configuration that creates pure bending.

Replaced the 12 kN·m option with 3 kN·m, the value produced by using the fixed-end coefficient PL/8 on a single 4 kN load instead of summing both loads' contributions on a simply supported span; the key of 8 kN·m re-derived cleanly and now sits at the top of the ascending list rather than in the middle.

Correct answer: (D)

Step 1 – Identify given information:

An open U-tube manometer uses mercury. The mercury surface on the tank side sits \(50\,\mathrm{mm}\) lower than on the open (atmospheric) side, so the vertical level difference is \(\Delta h = 50\,\mathrm{mm} = 0.050\,\mathrm{m}\). Mercury density is \(\rho_{\mathrm{Hg}} = 13{,}600\,\mathrm{kg/m}^{3}\) and \(g = 9.81\,\mathrm{m/s}^{2}\).

Step 2 – Apply the manometer relation:

For an open manometer, the gage pressure at the tank equals the weight of the mercury column corresponding to the level difference between the two free surfaces:

$$p_{\mathrm{gage}} = \rho_{\mathrm{Hg}}\, g\, \Delta h$$

The stated \(50\,\mathrm{mm}\) is already the net head; the two legs are not summed. Because the mercury is lower on the tank side, the tank pressure exceeds atmospheric, giving a positive gage pressure.

Step 3 – Substitute and solve:

$$p_{\mathrm{gage}} = 13{,}600 \times 9.81 \times 0.050 = 6{,}671\,\mathrm{Pa} \approx 6.7\,\mathrm{kPa}$$

Step 4 – Interpret the result:

The tank's gage pressure is about \(6.7\,\mathrm{kPa}\). Option (D) \(13.3\,\mathrm{kPa}\) would result only from doubling the head (treating each mercury leg separately), which is incorrect — a single \(50\,\mathrm{mm}\) difference already represents the full pressure head.

Correct answer: (C)

Step 1 – Eccentricity:

$$e = \frac{M}{P} = \frac{250}{600} = 0.417 \text{ m}$$

Step 2 – Pressure distribution:

$$q_{\text{avg}} = \frac{P}{A} = \frac{600}{3} = 200 \text{ kPa}$$

$$\frac{6e}{B} = \frac{6(0.417)}{3} = 0.833$$

$$q_{\max} = 200(1 + 0.833) = 367 \text{ kPa}$$

$$q_{\min} = 200(1 - 0.833) = 33 \text{ kPa}$$

Correct answer: (B)

Step 1 – Identify the given stress state:

Step 2 – Write the stress transformation equation (from the FE Handbook):

$$\sigma_\theta = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos(2\theta) + \tau_{xy}\sin(2\theta)$$

Step 3 – Evaluate the trigonometric terms at \( 2\theta = 2(45^\circ) = 90^\circ \):

$$\cos(90^\circ) = 0, \qquad \sin(90^\circ) = 1$$

Step 4 – Substitute all values into the equation:

$$\sigma_\theta = \frac{60 + 0}{2} + \frac{60 - 0}{2} \cdot 0 + 20 \cdot 1$$

$$\sigma_\theta = 30 + 0 + 20 = 50 \;\text{MPa}$$

Correct answer: (C)

Step 1 – Identify given information: A seepage flow net has streamlines and equipotential lines. Determine the shape of the resulting pattern.

Step 2 – Flow net construction rules: From the FE Handbook, a valid flow net must satisfy: (1) flow lines and equipotential lines are perpendicular, and (2) the resulting elements form curvilinear squares (equal width and height).

Step 3 – Evaluate options:

(A) Triangles would not satisfy the perpendicularity condition.

(B) Trapezoids would indicate unequal spacing.

(C) Curvilinear squares satisfy both conditions of the flow net.

(D) Rectangles would mean unequal head drop per element.

Step 4 – Answer: The correct shape is curvilinear squares, option (C). This property is what allows the simple seepage formula \( q = kH(N_f/N_d) \) to work.

Correct answer: (C)

Answer: B) 10

Integrate term by term:

$$\int(3x^2 + 2x - 1)\,dx = x^3 + x^2 - x + C$$

Evaluate from 0 to 2:

\(F(2) = 2^3 + 2^2 - 2 = 8 + 4 - 2 = 10\)

\(F(0) = 0^3 + 0^2 - 0 = 0\)

\(F(2) - F(0) = 10 - 0 = \textbf{10}\)

Answer: B) 3.2 MPa

Step 1: Calculate the mean:

$$\bar{x} = \frac{28 + 32 + 30 + 34 + 26}{5} = \frac{150}{5} = 30 \text{ MPa}$$

Step 2: Calculate the squared deviations:

\((28-30)^2 = 4\), \((32-30)^2 = 4\), \((30-30)^2 = 0\), \((34-30)^2 = 16\), \((26-30)^2 = 16\)

Step 3: Sum of squared deviations \(= 4 + 4 + 0 + 16 + 16 = 40\)

Step 4: Sample variance:

$$s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1} = \frac{40}{5-1} = \frac{40}{4} = 10$$

Step 5: Sample standard deviation:

$$s = \sqrt{10} = 3.16 \approx \textbf{3.2 MPa}$$

Answer: A) 8 kN

Take moments about point A \((\sum M_A = 0)\):

$$R_B \times 10 - 20 \times 4 = 0$$

$$R_B = \frac{20 \times 4}{10} = \frac{80}{10} = \textbf{8 kN}$$

Verification: \(R_A = 20 - 8 = 12\) kN. Check \(\sum M_B = 0\): \(12(10) - 20(6) = 120 - 120 = 0\). Confirmed.

Answer: B) 7.5 kN (Compression)

Step 1: By symmetry, \(R_A = R_C = \frac{12}{2} = 6\) kN (upward).

Step 2: Member AB goes from A(0,0) to B(3,4).

$$L_{AB} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \text{ m}$$

Direction cosines: \(\cos\theta = \frac{3}{5}\), \(\sin\theta = \frac{4}{5}\)

Step 3: At joint A, assuming all members in tension (forces directed away from joint):

$$\sum F_y = 0: \quad 6 + F_{AB}\left(\frac{4}{5}\right) = 0 \implies F_{AB} = -7.5 \text{ kN}$$

The negative sign means the assumed tension direction was wrong -- the member is in compression.

\(|F_{AB}| = \textbf{7.5 kN (Compression)}\)

Answer: C) 88.3 kN

The hydrostatic force on a submerged plane surface is:

$$F = \gamma \cdot \bar{h} \cdot A$$

where \(\bar{h}\) is the depth to the centroid of the gate.

The gate is 3 m tall with its top at the surface, so the centroid is at:

$$\bar{h} = \frac{3}{2} = 1.5 \text{ m below the surface}$$

Area: \(A = 2 \times 3 = 6 \text{ m}^2\)

$$F = 9.81 \times 1.5 \times 6 = \textbf{88.29} \approx \textbf{88.3 kN}$$

Answer: B) 183 kPa

Step 1: Find velocity in the narrow section using continuity \((A_1 V_1 = A_2 V_2)\):

The diameter ratio is \(\frac{200}{100} = 2\), so the area ratio is \(2^2 = 4\).

$$V_2 = V_1 \times \frac{A_1}{A_2} = 1.5 \times 4 = 6.0 \text{ m/s}$$

Step 2: Apply Bernoulli's equation (horizontal pipe, \(z_1 = z_2\)):

$$P_1 + \tfrac{1}{2}\rho V_1^2 = P_2 + \tfrac{1}{2}\rho V_2^2$$

$$200{,}000 + \tfrac{1}{2}(1000)(1.5)^2 = P_2 + \tfrac{1}{2}(1000)(6.0)^2$$

$$200{,}000 + 1{,}125 = P_2 + 18{,}000$$

$$P_2 = 201{,}125 - 18{,}000 = 183{,}125 \text{ Pa} \approx \textbf{183 kPa}$$

Answer: C) 40 kN·m

For a simply supported beam with a UDL, the maximum bending moment occurs at midspan:

$$M_{max} = \frac{wL^2}{8}$$

Where \(w = 5\) kN/m and \(L = 8\) m:

$$M_{max} = \frac{5 \times 8^2}{8} = \frac{5 \times 64}{8} = \frac{320}{8} = \textbf{40 kN}{\cdot}\textbf{m}$$

Answer: C) 617 kN

The Euler critical buckling load for a pin-pin column (\(K = 1.0\)) is:

$$P_{cr} = \frac{\pi^2 EI}{(KL)^2}$$

Convert units: \(E = 200{,}000\) N/mm\(^2\), \(I = 5.0 \times 10^6\) mm\(^4\), \(KL = 4{,}000\) mm

$$P_{cr} = \frac{\pi^2 \times 200{,}000 \times 5{,}000{,}000}{(4{,}000)^2}$$

$$P_{cr} = \frac{9.8696 \times 10^{12}}{16 \times 10^6} = 616{,}850 \text{ N} \approx \textbf{617 kN}$$

Answer: B) 101.9 kPa

Step 1: Total vertical stress at the bottom of the clay:

$$\sigma_{total} = \gamma_d \times H_{sand} + \gamma_{sat} \times H_{clay}$$

$$\sigma_{total} = 17 \times 3 + 20 \times 5 = 51 + 100 = 151 \text{ kPa}$$

Step 2: Pore water pressure at the bottom of the clay:

$$u = \gamma_w \times H_w = 9.81 \times 5 = 49.05 \text{ kPa}$$

(The water table is at the top of the clay, so the water column height equals the clay thickness.)

Step 3: Effective stress:

$$\sigma' = \sigma_{total} - u = 151 - 49.05 = \textbf{101.95} \approx \textbf{101.9 kPa}$$

Answer: A) CL

Step 1: Since more than 50% passes the No. 200 sieve (58%), the soil is fine-grained.

Step 2: Calculate the Plasticity Index (PI):

$$PI = LL - PL = 45 - 22 = 23$$

Step 3: Use the Plasticity Chart:

\(LL = 45\) (less than 50 → "L" suffix for low plasticity).

The A-line equation:

$$PI = 0.73 \times (LL - 20) = 0.73 \times 25 = 18.25$$

Since \(PI = 23 > 18.25\), the point plots above the A-line, indicating Clay (C).

Classification: CL (Low-plasticity clay)

Answer: B) $13,382

Use the future value formula for compound interest:

$$F = P(1 + i)^n$$

Where \(P = \$10{,}000\), \(i = 0.06\), \(n = 5\):

$$F = 10{,}000 \times (1.06)^5$$

$$(1.06)^5 = 1.33823$$

$$F = 10{,}000 \times 1.33823 = \textbf{\$13{,}382}$$

Answer: C) 101.9 MPa

Normal stress is defined as:

$$\sigma = \frac{P}{A}$$

Calculate the cross-sectional area:

$$A = \frac{\pi d^2}{4} = \frac{\pi (25)^2}{4} = \frac{\pi (625)}{4} = 490.87 \text{ mm}^2$$

Calculate stress:

$$\sigma = \frac{50{,}000}{490.87} = \textbf{101.9 MPa}$$